Notes of Bernard Maurey’s lecture nr 2

1. More examples of type and cotype

Let ${X}$ be a Banach space. The possible types are in ${I_T =[1,p_X)}$ or ${I_T =[1,p_X]}$, ${pX \leq 2}$. The possible cotypes are in ${I_C =[q_X ,\infty)}$ or ${I_C =(q_X ,\infty)}$, ${q_X \geq 2}$.

Cotype does not pass to quotients. Indeed, ${\ell_1}$ has cotype ${2}$, and every separable Banach space is a quotient of ${\ell_1}$. For instance, ${c_0}$ has trivial cotype.

The type of ${L_p (X)}$ is ${\min\{type(L_p),type(X)\}}$.

1.1. Complex interpolation

Type ${p}$ is equivalent to the map ${\ell_p (X)\rightarrow L_p (X)}$, ${(x_i)\mapsto \sum\epsilon_i x_i}$ being bounded. By interpolation, if ${X_0}$ has type ${p_0}$ and ${X_1}$ has type ${p_1}$, the interpolation space ${X_{\theta}}$, ${\theta\in (0,1)}$, has type ${p}$ such that

$\displaystyle \begin{array}{rcl} \frac{1}{p}=\frac{1-\theta}{p_0}+\frac{\theta}{p_1}. \end{array}$

Cotype does not work so nicely. It is expressible in terms of a map ${Rad(X)\rightarrow\ell_q (X)}$, where

$\displaystyle \begin{array}{rcl} Rad(X)=\{\sum\epsilon_i x_i \,;\,x_i \in X,\,\epsilon_i \in\{\pm 1\}\}\}\subset L_q (X). \end{array}$

If ${Rad(X)}$ is complemented, i.e. is the image of a projector ${L_q (X)\rightarrow Rad(X)}$ independent on ${q}$ (this is the notion of ${K}$-convexity), interpolation works for cotype.

1.2. Super

Clearly, type and cotype are super properties. A stronger property holds.

Definition 1 Say ${X}$ is crudely finitely representable in ${Y}$ if there exists ${C}$ such that every finite dimensional subspace of ${X}$ embeds with distorsion ${\leq C}$ in ${Y}$.

The same proof as for ordinary finite representability gives

Proposition 2 ${X}$ is crudely finitely representable in ${Y}$ iff ${X}$ is isomorphic to a subspace of an ultrapower ${Y^I /\mathcal{u}}$.

1.3. Type and finite representability

Theorem 3 (Pisier, Maurey-Pisier 1973) Let ${X}$ be a Banach space. Then ${X}$ has trivial type iff ${\ell_1}$ is finitely representable in ${X}$. ${X}$ has trivial cotype iff ${c_0}$ is finitely representable in ${X}$.

Lemma 4 Let ${Y_1 ,\ldots,Y_n}$ be independent ${X}$-valued random variables with a symmetric distribution. Let ${\epsilon_i}$ be independent Bernoulli variables. Then ${(\epsilon_1 Y_1 ,\ldots,\epsilon_n Y_n)}$ and ${Y_1 ,\ldots,Y_n}$ have the same distribution.

Proof: of Theorem 3. Let ${a_n}$ denote the best constant in the inequality

$\displaystyle \begin{array}{rcl} (\mathop{\mathbb E}|\sum_{i=1}^{n}\epsilon_i x_i|^2 )^{1/2} \leq \sqrt{n}a_n (\sum|x_i|^2 )^{1/2} . \end{array}$

Then ${a_n \leq 1}$. Let us show that ${(a_n)}$ is submultiplicative, i.e. ${a_{mn}\leq a_m a_n}$. Indeed, group ${mn}$ variables in ${m}$ packages of ${n}$.

Assume that ${a_n =1}$ for all ${n}$. So for every ${n}$, there exist points ${\mathop{\mathbb E}|\sum_{i=1}^{n}\epsilon_i x_i|^2}$ is close to ${a_n^2 (\sum|x_i|^2 )}$. Go to an ultrapower to find vectors ${\xi_i}$ such that ${\mathop{\mathbb E}|\sum_{i=1}^{n}\epsilon_i \xi_i|^2 =a_n^2 (\sum|\xi_i|^2 )}$. The equality case in the chain of inequalities leading to ${a_n \leq 1}$ shows that all norms ${|\xi_i |=1}$ and for all choices of signs, ${|\sum\pm \xi_i|=n}$. Fix signs ${\epsilon_i}$. If ${x^*}$ is a unit functional, ${x^*(\sum\epsilon_i \xi_i)=n}$, thus ${x^* (\xi_i)=\epsilon_i}$. In other words, these functionals give an isometry of span${(\x_i)}$ onto ${\ell_1}$.

Otherwise, there exists ${n_0}$ such that ${a_{n_0}<1}$. Then there exists ${C>0}$, ${\alpha>0}$, such that ${a_n \leq C n^{-\alpha}}$. Let ${p\geq p_X}$. Given a family ${x_i \in X}$, split it according to norms,

$\displaystyle \begin{array}{rcl} A_k =\{i\,;\,2^{-k-1}<|x_i|\leq 2^{-k}\}. \end{array}$

Then

$\displaystyle \begin{array}{rcl} (\mathop{\mathbb E}|\sum_{i\in A_k}\epsilon_i x_i|^2 )^{1/2} \leq \sqrt{|A_k|}a_{|A_k|}2^{-k}\sqrt{|A_k|}. \end{array}$

Summing over ${k}$ gives the wanted estimate. $\Box$

1.4. Ribe’s Theorem

It is our first nonlinear result.

Theorem 5 (Ribe 1976) If ${T:X\rightarrow Y}$ is a uniform homeomorphism, then ${X}$ is crudely finitely representable in ${Y}$ and ${Y}$ is crudely finitely representable in ${X}$.

Proof: Upgrade the uniform homeomorphism to a bi-Lipschitz map. Use differentiability on finite dimensional subspaces to improve bi-Lipschitz maps to isomorphisms. $\Box$

This implies that ${L_p [0,1]}$ and ${L_q [0,1]}$ are not uniformly homeomorphic, an earlier result due to Enflo. Enflo’s proof relies on a metric notion of type, on which Naor and Schechtman, Mendel and Naor have elaborated since.

Definition 6 Let ${E}$ be a metric space. Say ${E}$ has Enflo type ${p}$ if there is a constant ${C>0}$ such that for all ${f:\{\pm 1\}^n\rightarrow E}$ ,

$\displaystyle \begin{array}{rcl} \mathop{\mathbb E}_{\epsilon}d(f(\epsilon),f(-\epsilon))^p \leq C^p \sum_{j=1}^{n}\mathop{\mathbb E}_{\epsilon}d(f(\epsilon),f(R_j \epsilon))^p , \end{array}$

where ${R_j}$ changes the sign of the ${j}$-th coordinate.

Remark 1 For a Banach space, Enflo type ${p}$ implies usual type ${p}$.

1.5. Proof of Ribe’s Theorem

Let ${T:X\rightarrow Y}$ be a uniform homeomorphism. Then ${T}$ is large-scale Lipschitz: ${|T x_1 -T x_2|\leq C|x_1 -x_2|}$ provided ${|x_1 -x_2|>1}$. Go to an ultra product to get rid the the large-scale restriction. This means setting ${\tilde{T}x =(T(kx)/k)_{k\in{\mathbb N}}\in Y^{{\mathbb N}}}$. This is bi-Lischitz: ${X\rightarrow Y^{{\mathbb N}}_{\mathcal{U}}}$. On every finite dimensional subspace ${X_0 \subset X}$, ${\tilde{T}}$ has points of differentiability. The differential is a linear isomorphism of ${X_0}$ onto a linear subspace of ${Y^{{\mathbb N}}_{\mathcal{U}}}$. So ${X}$ is coarsely finitely representable in ${Y^{{\mathbb N}}_{\mathcal{U}}}$, and thus in ${Y}$.

2. ${B}$-convexity

This goes back to Beck and Giesy. They were lookibg for a law of large numbers for Banach space valued random variables.

Lemma 7 Let ${B}$ have type ${p>1}$. Let ${X_1 ,\ldots,X_n ,\ldots}$ be independent centered ${B}$-valued random variables. Then

$\displaystyle \begin{array}{rcl} \mathop{\mathbb E}|\sum_{i=1}^{n}X_i|^p \leq 2^{p}T_{p}^{p}\sum_{i=1}^{n}\mathop{\mathbb E}|X_i |^p. \end{array}$

Proof: Reduce to the case of symmetric distributions by replacing ${X_i}$ by ${X_i -X'_i}$ where ${X'_i}$ is an independent copy of ${X_i}$. This produces the factor ${2^p}$. Then apply Jensen’s inequality

$\displaystyle \begin{array}{rcl} \mathop{\mathbb E}|\sum_{i=1}^{n}X_i|^p \leq\mathop{\mathbb E}|\sum_{i=1}^{n}X_i|^p \end{array}$

$\Box$

It implies that

$\displaystyle \begin{array}{rcl} \mathop{\mathbb E}|\frac{1}{n}\sum_{i=1}^{n}X_i|\leq(\mathop{\mathbb E}|\frac{1}{n}\sum_{i=1}^{n}X_i|^p)^{1/p}\leq 2T_p n^{(1-p)/p} ,\end{array}$

which tends to ${0}$, this is a law of large numbers.

2.1. Application: approximation of convex hulls

Corollary 8 Let ${B}$ be a Banach space with type ${p>1}$. Let ${A\subset U_B}$. Let ${x\in Conv(A)}$. There exist ${a_1 ,\ldots,a_n \in A}$ such that

$\displaystyle \begin{array}{rcl} |s-\frac{1}{n}\sum_{i=1}^{n}a_j|\leq 4 T_p n^{(1-p)/p}. \end{array}$

Proof: Apply the law of large numbers to independent copies of the following random variable ${U}$. By assumption, there exist ${\lambda_i \geq 0}$, ${\sum\lambda_i =1}$ such that ${x=\sum\lambda_i x_i}$. So let ${U=x_i}$ with probability ${\lambda_i}$. $\Box$

3. Factorization

The following was one of the first results on isomorphism types.

Theorem 9 (Kwapién) If a Banach space has simultaneously type ${2}$ and cotype ${2}$, then is is isomorphic to a Hilbert space.

Up to now we have uses Bernoulli averages. Why not use Gaussian averages ? Grouping ${mn}$ Bernoulli variables (divided by ${n}$) in packages of ${n}$ produces ${m}$ variables which are approximately standard Gaussian (central limit theorem). So we can use Gaussian averages instead.

Lemma 10 Let ${G_1}$, ${G_2}$ be ${E}$-valued centered Gaussian vectors. Assume that for all ${t\in E^*}$,

$\displaystyle \begin{array}{rcl} \mathop{\mathbb E}(tG_1)^2 :=Q_1 (t)\leq Q_2 (t):=\mathop{\mathbb E}(tG_2)^2 . \end{array}$

Then for every convex function ${f}$,

$\displaystyle \begin{array}{rcl} \mathop{\mathbb E}(f(G_1))\leq \mathop{\mathbb E}(f(G_2)). \end{array}$

Proof: Pick a common orthogonal basis for the quadratic forms ${Q_1}$ and ${Q_2}$. Use such a basis to find ${G'}$ independant from ${G_1}$ such that both ${G_1 \pm G'}$ have the same distribution as ${G_1}$. $\Box$

Let ${E}$ be a finite dimensional normed space. Say a homogeneous function ${\phi:E\rightarrow{\mathbb R}_+}$ is a cotype ${2}$ function if for all ${x_1 ,\ldots,x_n \in E}$,

$\displaystyle \begin{array}{rcl} \sum\phi(x_i)^2 \leq C_{\phi}^2 \mathop{\mathbb E}|\sum g_i x_i|^2 \end{array}$

where ${g_i}$ are independent standard Gaussians.

Lemma 11 There exists a measure on the unit sphere ${S}$ of ${S^*}$ such that for all ${x\in E}$,

$\displaystyle \begin{array}{rcl} \frac{1}{C_{\phi}}\phi(x)\leq(\int_{S}(t\cdot x)^2 \,d\mu(t))^{1/2}\leq T_{2,E}|x|. \end{array}$

Proof: Set ${f_x (t)=\sum(t\cdot x_i)^2}$,

$\displaystyle \begin{array}{rcl} C_1=\{f_x -g \,;\, \sum|x_i|^20\textrm{ on }S\}, \end{array}$

$\displaystyle \begin{array}{rcl} C_2=\{f_y \,;\,\sum|y_i|^2 \geq C_{\phi}^{2}\}. \end{array}$

These are disjoint convex sets $\Box$

I have to run to catch the 4:30 train to Orsay…